In this lecture

1
In this lecture

• Number Theory
• ? Quotient and Remainder
• ? Floor and Ceiling
• Proofs
• ? Direct proofs and Counterexamples
  (cont.)
• ? Division into cases

2
Quotient-Remainder Theorem

• Theorem For ? n?Z and d?Z
• ?! q,r?Z such that
• ndqr and 0rltd.
• q is called quotient r is called remainder.
• Notation q n div d r n mod d.
• Examples 1) 53 865. Hence
• 53 div 8 6 53 mod 8 5.
• 2) -29 7(-5)6. Hence
• -29 div 7 -5 -29 mod 7 6.

3
Example of using div and mod

• Last year Halloween was on Sunday.
• Q. What day is Halloween this year?
• Solution There are 365 days between
• 10/31/10 and 10/31/11.
• 365 mod 7 1.
• Thus, if 10/31/10 is Sunday
• then 10/31/11 is Monday.

4
Proof Technique Division into Cases

• Suppose at some stage of a proof
• ? we know that
• A1 or A2 or A3 or or An is true
• ? want to deduce a conclusion C.
• Use division into cases
• Show A1?C, A2?C, , An?C.
• Conclude that C is true.

5
Division into Cases Example

• Proposition If n?Z s.t.
• neither of 2 or 3 divide n, (1)
• then n2 mod 12 1. (2)
• Proof Suppose n?Z s.t. neither of 2 or 3 divide
  n.
• By quotient-remainder theorem,
• exactly one of the following is true
• a) n6k, b) n6k1, c) n6k2, d) n6k3,
• e) n6k4, f) n6k5 for some integer
  k. (3)
• n cant be 6k, 6k2 or 6k4 because
• in that case 2 n (which contradicts (1) ).
  (4)
• n cant be 6k3 because in that case 3 n
• (which contradicts (1) ). (5)

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Division into Cases Example(cont.)

• Proof(cont.) Based on (3), (4) and (5),
• either n6k1 or n6k5.
• Lets show (2) for each of these two cases.
• Case 1 Suppose n6k1.
• Then n2 (6k1)236k212k1 (by basic
  algebra)
• 12(3k2k)1 (6)
• Let p3k2k. Then p is an integer.
• n2 12p1 . ( by substitution in (6) )
• Hence n2 mod 12 1 by quotient-remainder
  th-m.
• Case 2 Suppose n6k5. (exercise)
  

7
Floor and Ceiling

• Definition For any real number x,
• ? the floor of x
• the unique integer n s.t. n x lt n1
• ? the ceiling of x
• the unique integer n s.t. n-1 lt x n.
• Examples

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Properties of Floor and Ceiling

• For ? x?R and ?m?Z , .
• 
  is false.
• Counterexample For x1.7, y2.8,
• Note If x,ygt0 and
• the sum of their fractional parts is lt1 then

9
Properties of Floor and Ceiling

• Theorem For ? n?Z ,
• Proof Case 1 Suppose n is odd.
• Then n2k1 for some integer k. (1)

10
Properties of Floor and Ceiling

• Proof (cont.) By substitution from (1),
• (2)
• because k?Z and k k1/2 lt k1.
• On the other hand,
• n2k1 ? k(n-1)/2. (by basic algebra) (3)
• Based on (2) and (3),
• Case 2 n is even (left as exercise).